NC54834. 大强的奋斗之路
描述
农民大强经过多年奋斗终于攒下了m块钱,可以购买属于自己的土地。土地有n种,第i种的价格为ai/m2,年收益为bi/m2。每年年初,大强可以决定是否购买土地;所有的土地都在年末给他带来收入。但是年收入太高容易引起他人的嫉妒,因此大强会Q次问你,在年收益不超过ci的情况下,攒足至少di块钱至少要几年。
输入描述
第1行包括三个正整数n,m,Q。
第2行到第n+1行,每行包括两个正整数ai,bi。
接下来Q行,每行包括两个正整数ci,di。
输出描述
输出Q行,每行一个整数,表示每次询问的答案。若永远无法达到,输出-1。
示例1
输入:
3 3 2 2 1 3 2 6 8 3 9 6 100
输出:
4 20
说明:
对于第一组询问,在第一年年初购进1m2第二种土地,在第二年年初购进1m2第一种土地。此时大强的年收益为3块钱,手中没有钱。到第四年末刚好攒满9块钱。C++14(g++5.4) 解法, 执行用时: 373ms, 内存消耗: 3680K, 提交时间: 2020-01-17 21:44:40
#include<bits/stdc++.h> typedef unsigned int uint; typedef long long ll; typedef unsigned long long ull; typedef double lf; typedef long double llf; typedef std::pair<int,int> pii; #define xx first #define yy second template<typename T> inline T max(T a,T b){return a>b?a:b;} template<typename T> inline T min(T a,T b){return a<b?a:b;} template<typename T> inline T abs(T a){return a>0?a:-a;} template<typename T> inline bool repr(T &a,T b){return a<b?a=b,1:0;} template<typename T> inline bool repl(T &a,T b){return a>b?a=b,1:0;} template<typename T> inline T gcd(T a,T b){T t;if(a<b){while(a){t=a;a=b%a;b=t;}return b;}else{while(b){t=b;b=a%b;a=t;}return a;}} template<typename T> inline T sqr(T x){return x*x;} #define mp(a,b) std::make_pair(a,b) #define pb push_back #define I __attribute__((always_inline))inline #define mset(a,b) memset(a,b,sizeof(a)) #define mcpy(a,b) memcpy(a,b,sizeof(a)) #define fo0(i,n) for(int i=0,i##end=n;i<i##end;i++) #define fo1(i,n) for(int i=1,i##end=n;i<=i##end;i++) #define fo(i,a,b) for(int i=a,i##end=b;i<=i##end;i++) #define fd0(i,n) for(int i=(n)-1;~i;i--) #define fd1(i,n) for(int i=n;i;i--) #define fd(i,a,b) for(int i=a,i##end=b;i>=i##end;i--) #define foe(i,x)for(__typeof((x).end())i=(x).begin();i!=(x).end();++i) #define fre(i,x)for(__typeof((x).rend())i=(x).rbegin();i!=(x).rend();++i) struct Cg{I char operator()(){return getchar();}}; struct Cp{I void operator()(char x){putchar(x);}}; #define OP operator #define RT return *this; #define UC unsigned char #define RX x=0;UC t=P();while((t<'0'||t>'9')&&t!='-')t=P();bool f=0;\ if(t=='-')t=P(),f=1;x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0' #define RL if(t=='.'){lf u=0.1;for(t=P();t>='0'&&t<='9';t=P(),u*=0.1)x+=u*(t-'0');}if(f)x=-x #define RU x=0;UC t=P();while(t<'0'||t>'9')t=P();x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0' #define TR *this,x;return x; I bool IS(char x){return x==10||x==13||x==' ';}template<typename T>struct Fr{T P;I Fr&OP,(int&x) {RX;if(f)x=-x;RT}I OP int(){int x;TR}I Fr&OP,(ll &x){RX;if(f)x=-x;RT}I OP ll(){ll x;TR}I Fr&OP,(char&x) {for(x=P();IS(x);x=P());RT}I OP char(){char x;TR}I Fr&OP,(char*x){char t=P();for(;IS(t);t=P());if(~t){for(;!IS (t)&&~t;t=P())*x++=t;}*x++=0;RT}I Fr&OP,(lf&x){RX;RL;RT}I OP lf(){lf x;TR}I Fr&OP,(llf&x){RX;RL;RT}I OP llf() {llf x;TR}I Fr&OP,(uint&x){RU;RT}I OP uint(){uint x;TR}I Fr&OP,(ull&x){RU;RT}I OP ull(){ull x;TR}};Fr<Cg>in; #define WI(S) if(x){if(x<0)P('-'),x=-x;UC s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0') #define WL if(y){lf t=0.5;for(int i=y;i--;)t*=0.1;if(x>=0)x+=t;else x-=t,P('-');*this,(ll)(abs(x));P('.');if(x<0)\ x=-x;while(y--){x*=10;x-=floor(x*0.1)*10;P(((int)x)%10+'0');}}else if(x>=0)*this,(ll)(x+0.5);else *this,(ll)(x-0.5); #define WU(S) if(x){UC s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0') template<typename T>struct Fw{T P;I Fw&OP,(int x){WI(10);RT}I Fw&OP()(int x){WI(10);RT}I Fw&OP,(uint x){WU(10);RT} I Fw&OP()(uint x){WU(10);RT}I Fw&OP,(ll x){WI(19);RT}I Fw&OP()(ll x){WI(19);RT}I Fw&OP,(ull x){WU(20);RT}I Fw&OP() (ull x){WU(20);RT}I Fw&OP,(char x){P(x);RT}I Fw&OP()(char x){P(x);RT}I Fw&OP,(const char*x){while(*x)P(*x++);RT} I Fw&OP()(const char*x){while(*x)P(*x++);RT}I Fw&OP()(lf x,int y){WL;RT}I Fw&OP()(llf x,int y){WL;RT}};Fw<Cp>out; const int N=1007,M=10007,Q=100007; std::vector<pii>qs[M]; int n,m,cm,q,a[N],b[N],f[M],qi[M],qa[M],c[Q],d[Q],ans[Q]; pii s[N]; int main() { //freopen("in.txt","r",stdin); in,n,m,q; fo1(i,n)in,s[i].xx,s[i].yy; std::sort(s+1,s+n+1); fo1(i,n)a[i]=s[i].xx,b[i]=s[i].yy; //fo1(i,n)in,a[i],b[i]; fo1(i,q)in,c[i],d[i]; fo1(i,q)repr(cm,c[i]),qs[c[i]].pb(mp(d[i],i)); //out,'/',cm,'\n'; fo(i,0,cm)f[i]=-1e9; f[0]=m; fo1(i,cm)std::sort(qs[i].begin(),qs[i].end()); fo1(i,q)ans[i]=-1; fo(i,0,cm)qa[i]=1; //out,qs[0].size(),'\n'; int rem=q; fo(j,0,cm) { int&qt=qi[j]; //out,'/',j,' ',qt,' ',qs[j].size(),'\n'; for(;qt<qs[j].size()&&qs[j][qt].xx<=m;qt++) ans[qs[j][qt].yy]=0,rem--; } fo1(T,M) { //fo(j,0,cm)fo1(i,n)if(f[j]>=a[i]&&j+b[i]<=cm)repr(f[j+b[i]],f[j]-a[i]); fo(j,0,cm) { int&qt=qa[j]; for(;qt<=n&&a[qt]<=f[j];qt++) if(j+b[qt]<=cm)repr(f[j+b[qt]],f[j]-a[qt]); } int mxf=0; fo(j,0,cm) { f[j]+=j; repr(mxf,f[j]); int&qt=qi[j]; //out,'/',j,' ',qt,' ',qs[j].size(),'\n'; for(;qt<qs[j].size()&&qs[j][qt].xx<=mxf;qt++) ans[qs[j][qt].yy]=T,rem--; } if(!rem)break; } fo1(i,q)out,ans[i],'\n'; }
C++(clang++ 11.0.1) 解法, 执行用时: 76ms, 内存消耗: 2180K, 提交时间: 2022-11-19 21:37:14
#include<cstdio> #include<algorithm> using namespace std; const int N=200010; int n,m,nn,x,top,a[N],b[N],f[N],s[N],ans[N]; struct que {int lim,aim,id;} c[N]; struct poi {double x,y;} p[N]; bool operator < (que a,que b) {return a.lim<b.lim;} inline int read() { int tmp=0, fh=1; char c=getchar(); while (c<'0'||c>'9') {if (c=='-') fh=-1; c=getchar();} while (c>='0'&&c<='9') tmp=tmp*10+c-48, c=getchar(); return tmp*fh; } inline bool below (poi a,int l) {return a.y<=a.x*l+f[l];} inline poi jiao(int l1,int l2) { double x=1.0*(f[l2]-f[l1])/(l1-l2); double y=l1*x+f[l1]; return (poi){x,y}; } int main() { n=read(); f[0]=read(); m=read(); for (int i=1;i<=n;i++) a[i]=read(), b[i]=read(); for (int i=1;i<=m;i++) { c[i].lim=read(); c[i].aim=read(); c[i].id=i; nn=max(nn,c[i].lim); } for (int i=1;i<=m;i++) ans[i]=1e9; sort(c+1,c+m+1); for (int i=1;i<=nn;i++) f[i]=-1e9; for (int i=1;i<=n;i++) if (a[i]<=f[0]) f[b[i]]=max(f[b[i]],f[0]-a[i]); //for (int i=0;i<=nn;i++) printf("%d\n",f[i]); for (int i=1,k=1;i<=nn;i++) { if (f[i]>-1e9) { while (top>1&&below(p[top-1],i)) top--; if (top) p[top]=jiao(s[top],i); s[++top]=i; } while (k<=m&&c[k].lim==i) { //printf("QAQ %d\n",c[k].id); if (c[k].aim<=f[0]) {ans[c[k].id]=0; k++; continue;} if (!top) {k++; continue;} //printf("QAQ %d %d\n",c[k].id,f[i]); int l,r,mid; l=0; r=top; while (l+1<r) { mid=(l+r)>>1; if (p[mid].y>=c[k].aim) r=mid; else l=mid; } //printf(" %d %d\n",r,s[r]); x=(c[k].aim-f[s[r]])/s[r]; if (s[r]*x+f[s[r]]<c[k].aim) x++; ans[c[k].id]=x; k++; } if (f[i]>-1e9) { for (int j=1;j<=n;j++) if (i+b[j]<=nn) { x=(i==0?0:max((a[j]-f[i])/i,0)); if (i*x+f[i]<a[j]) x++; f[i+b[j]]=max(f[i+b[j]],f[i]-a[j]-b[j]*x); } } } //for (int i=0;i<=nn;i++) printf("%d %d\n",i,f[i]); puts(""); for (int i=1;i<=m;i++) printf("%d\n",ans[i]<1e9?ans[i]:-1); return 0; }